Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

思路:典型的动态规划问题。result[i][j]表示(0,0)到(i,j)的最小值。result[i][j]=min(result[i][j-1],result[i-1][j])+grid[i][j];

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        int m=grid.size();
        int n=grid[0].size();
        if(m==0||n==0)
            return 0;
        int result[m][n];
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i==0)
                {
                    if(j==0)
                        result[i][j]=grid[i][j];
                    else
                        result[i][j]=result[i][j-1]+grid[i][j];
                }
                else
                {
                    if(j==0)
                        result[i][j]=result[i-1][j]+grid[i][j];
                    else
                        result[i][j]=min(result[i-1][j],result[i][j-1])+grid[i][j];
                }
            }
        }
        return result[m-1][n-1];
    }
};
原文地址:https://www.cnblogs.com/awy-blog/p/3619800.html